Physics 361
Principles of Modern Physics
Lecture 3
Blackbody Radiation
And Its Implications on Quantum
Properties of Light
How Does Blackbody Radiation
Form?
Go back to an ideal gas.
All molecules move
randomly.
Each one has on
average a kinetic
3
energy of 2 kT .
1
kT
2
for each mode
of energy storage.
“Degree of
freedom.”
The higher the T,
the greater the
movement of
molecules.
How Does Blackbody Radiation
Form?
Go back to an ideal gas.
Since all matter is made
of charged particles, this
random movement of
charge should emit
electromagnetic waves.
The higher the
temperature, the more
energy is radiated from
the matter.
Remember our earlier
arguments about E&M
wave emitters.
EM Waves by an Antenna
• Two rods are connected to an ac source, charges oscillate between the
rods (a)
• As oscillations continue, the rods become less charged, the field near
the charges decreases and the field produced at t = 0 moves away from
the rod (b)
• The charges and field reverse (c)
• The oscillations continue (d)
So everything at a finite
temperature emits light.
But why do we call them “black
bodies”??
Consider Two Identical Materials at
the Same Temperature
They are in thermal equilibrium, so the net energy across dashed line must be zero!
Now add a perfectly reflecting
coating on only one of them.
In equilibrium the net energy across dashed line must still be zero! So the
perfectly reflecting material does not emit radiation!!
Thus, perfectly reflecting bodies do
not absorb or emit light.
Reflecting bodies have zero “emissitivity” or “absorptivity” – the same
thing.
Something that emits a lot of light is
not reflective – thus it is black!!
Blackbody Radiation – Opening
in a cavity is a good
approximation
Emitted light from opening should
be proportional to average
energy density within the
enclosed cavity.
Energy density per
small increment of
wavelength is .
In classical wave picture of light,
the number of wave excitations
(modes of oscillation) grows as
an inverse power of wavelength.
That is – the smaller the
wavelength, the more types of
oscillations you can fit in an
enclosure.
Like fitting marbles in a jar. The
smaller the marbles, the more
you can fit into a jar.
For wave modes, this is the same.
The smaller their wavelength, the
more of their mode you can fit
into an enclosed space.
Mode density per
small wavelength
increment works out
to be
8p
n(l ) = 4
l
*Note: we will work
out details of mode
counting later in this
course.
So we expect more small
wavelength oscillations inside the
cavity. How much energy is
contained in each mode?
Classical Equipartition Theorem
Stick an oscillator in container
(assume motion only allowed
in 1D). How much energy is
on average stored in the
oscillator?
1
kT
2
must be stored in kinetic
energy due to collisions.
Since oscillators vary
periodically in having energy
either all in kinetic or all in
potential, we expect another
to be stored on average in
potential.
1
kT
2
Thus, a total of kT should be
stored in an oscillator mode in
thermal equilibrium.
Total energy density per unit
wavelength is then
u(l ) = kTn(l ) = kT
8p
l4
This diverges as wavelength goes to zero!!
If classical blackbody radiation
(“Rayleigh-Jeans” behavior) were
correct... you’d need a really good
pair of sunglasses.
• Everything would emit an infinite
amount of radiation (and
energy).
• Raises important questions...
• Where would objects store this
infinite energy source?
• Wouldn’t the world be a very
bright place??
The Ultraviolet Catastrophe
• Classical theory does not match
the experimental data
• At long wavelengths, the match
is good
• At short wavelengths, classical
theory predicts infinite energy
density
• At short wavelengths,
experiment shows approximately
zero energy density
• This contradiction is called the
ultraviolet catastrophe
Experimental Blackbody
Radiation Graph
•
•
Experimental data for distribution of
energy in blackbody radiation at right.
As the temperature increases, the total
amount of energy and emitted intensity
increases
– Shown by the area under the curve

•
As the temperature increases, the
peak of the distribution shifts to
shorter wavelengths (“Wien’s
displacement law”)
λmax T = 0.2898 x 10-2 m • K
Total emitted intensity is proportional to
T 4.
Planck’s Resolution
•
•
•
•
•
Planck hypothesized that the blackbody radiation was produced by
resonators
– Resonators were submicroscopic charged oscillators
The resonators could only have discrete energies
– En = n h ƒ
•
•
•
n is called the quantum number
ƒ is the frequency of vibration
h is Planck’s constant, 6.626 x 10-34 J s
Key point is quantized energy states
Mathematically correct, but he did not explain why this quantization occurs
or specifically where it occurs.
Gives a correct description of blackbody radiation.
Modern View – The
Electromagnetic Field is Quantized
into
Photons
• The E&M waves have only allowed discrete allowed
numbers
– En = n h ƒ
• n is called the quantum number
• ƒ is the frequency of vibration
• h is Planck’s constant, 6.626 x 10-34 J s
• These quantum numbers are the actual number of E&M
radiation particles (called “photons”) present in a mode.
• Since small wavelength excitations have very large
energies, they would be unlikely to be thermally excited
and there would not be kT of energy for each of these
high-energy modes. This is the source of the exponential
term in Plank’s description. (This can be derived using
quantum statistics later in the course.)
Conclusions about E&M radiation
• Not only is the energy dependent on the
frequency rather than the amplitude, but the
radiation acts like discrete particles that can be
counted.
• Next lecture we will explore collisions of these
photons with matter to see whether they actually
behave like particles.